Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Wednesday, September 5, 2018

### Geometry Problem 1385: Triangle, Orthocenter, Circle, Circumcircle, Angle Bisector, Midpoint, Collinear Points

Labels:
angle bisector,
circle,
circumcircle,
collinear,
geometry problem,
mathematics,
midpoint,
orthocenter,
proof,
triangle

Subscribe to:
Post Comments (Atom)

https://photos.app.goo.gl/Ki4AYSREXnaEaSAZ7

ReplyDeleteLet BD and BO extended meet circle O at P and Q

Let M is the midpoint of AC and NQ meet OP at M’

Let BM meet HO at R ( R is the centroid of triange ABC)

Note that HRO is the Euler line of triangle ABC

We have RO/RH= OM/BH= ½ (1) ( Property of Euler line)

Triangle BHQ simillar to OM’Q.. ( case AA)

So QO/QB= OM’/HB= ½ .. (2)

Compare (1) and (2) we have OM’=OM => M coincide to M’

So Q, H, N, M are collinear